I need to know how to go about finding the speed.(adsbygoogle = window.adsbygoogle || []).push({});

A block B of mass 5 kg is fastened to one end of each of two springs. The other ends of the springs are attached to fixed points A and C, 4 metres apart on a smooth horizontal surface, as shown in the diagram.

Spring AB has natural length 2 metres and modulus of elasticity 30 N, while BC has natural length 1 metre and modulus 40 N (you may assume that the springs meet at the centre of B).

If the block is moved 0.5 metres towards C from its equilibrium position and then released, determine its speed as it passes through its equilibrium position.

this is what i have tried

Energy in string when moved 0.5m = total mech energy before the 0.5 movement

E= 30*(0.5)^2/(2*2) = 1.575 J

KE= 1/2*5*v^2 = 5/2*v^2 J and

E in string 30*(3/8)^2/2*2 = 7.93 J

7.93 + 5/2v^2 = 1.875

this leads me to taking square root of a -ve number

could someone help me!

thanks in advance

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# Homework Help: Total Elastic Potential Energy 2

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